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Given that Ak=k(k1)2cosk(k1)π2, find A19+A20++A98.

 Aug 2, 2022
 #1
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k(k1)2 is really just the sum of integers from 1 to k-1, and the cosine of some integer multiple π oscillates between -1 and 1 (if the integer is even, it's 1, and if it's odd, it's -1). Here, the sum alternates 2 negatives and 2 positives (starting with k=19)

This means that the first 4 terms of the sum are

(1+2+3++18)(1+2+3++19)+(1+2+3++20)+(1+2+3++21)=20+19+20+21=4(20)

(notice a bunch of cancellation happens between the sums, as the 1+2+3+...18 is repeated throughout all the 4 terms)

So for every group of 4 terms, the sum of the group is the biggest number of the sum of the 3rd term of that group times 4.

Therefore, the sum is:

4(20+24+28++96)

Use the sum formula to obtain your answer.

 Aug 2, 2022
 #2
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+1

We want to compute cosk(k1)π2 for various intergers k. The first values are as follows:

 kcosk(k1)π21121314151617181

The values seem to be periodic with period 4. If we let f(k)=k(k1)π2, then 

f(k+4)f(k)=(k+4)(k+3)π2k(k1)π2=(4k+6)π=(2k+3)2π

Since f(k+4) and f(k) differ by a mutiple of 2πcosf(k) is indeed period with period of 4. So for a positive integer m, 

A4m1=(4m1)(4m2)2cosf(4m1)=(4m1)(4m2)2,A4m=(4m)(4m1)2cosf(4m)=(4m)(4m1)2,A4m+1=(4m+1)(4m)2cosf(4m+1)=(4m+1)(4m)2,A4m+2=(4m+2)(4m+1)2cosf(4m+2)=(4m+2)(4m+1)2,

Therefore:

A4m1+A4m+A4m+1+A4m+2=(4m1)(4m2)2+(4m)(4m1)2+(4m+1)(4m)2(4m+2)(4m+1)2=2.

Summing over 5m24, we find that A19+A20++A98=20(2)=40.

 Aug 3, 2022
 #3
avatar+506 
0

thanks guest, my answer is very far off lol

textot  Aug 3, 2022

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